Exercise 18.6.K in Vakil's Foundations of Algebraic geometry is:
Let $X$ be a projective scheme of dimension $\geq 1$ over a field $k$, with a fixed closed immersion $i : X \rightarrow \mathbb{P}^n_k$. Let $H= V(f)$ be a hypersurface not containing any associated points of $X$. Then, $\deg(H \cap X) = \deg(H) \cdot \deg(X)$
Theorem 1.7.7 in Hartshorne's Algebraic Geometry is : Let $Y$ be a variety of dimension $\geq 1$ in $\mathbb{P}^n_k$. Let $H$ be a hypersurface not containing $Y$. Let $Z_1, \dots, Z_s$ be the irreducible components of $Y \cap H$, corresponding to prime ideals $p_1, \dots, p_s$. Define $i(Y, H ; Z_j)$ to be the length of $k[x_0, \dots, x_n] / (I(Y) + I(H))$ at $p_j$. Then, $\sum_{j=1}^s i(Y, H ; Z_j) \cdot \deg(Z_j) = \deg(Y) \cdot \deg(H)$.
I want to use theorem 1.7.7 to solve exercise 18.6.K. The difficulty that I'm having is that in chapter 1 of Hartshorne, only integral varieties over algebraically closed fields are considered. But in Vakil, $X$ can be reduce or have multiple components.
My attempt at solving this problem is: if I take $X = \operatorname{Proj}(k[x_0, \dots, x_n] / I)$, I can then take a primary decomposition of $I = \cap Q_i$ to consider each individual irreducible component separately, therefore I reduce to where $X$ is irreducible. But then I'm not sure how to relate the degree of $k[x_0, \dots, x_n] / I$ with $k[x_0, \dots, x_n] / Q_i$. Furthermore, $Q_i$ is just primary , and not prime. So I can't apply theorem 1.7.7 directly. I also don't know how relate the degree of $k[x_0, \dots, x_n] / Q_i$ with its length over $p_i$.
Best Answer
Let me expand my comments in to an answer. Writing $S$ for the homogeneous coordinate ring of $X\subset\Bbb P^n$, $f$ for the defining equation of $H$, and $d=\deg f$, we have the exact sequence $$0\to S(-d)\stackrel{\cdot f}{\to} S\to S/(f)\to 0,$$ where the first map is injective because $V(f)$ contains no associated points of $X$. Now we can take Hilbert polynomials to see that $H_X(n)-H_X(n-d)=H_{X\cap H}(n)$. Writing $H_X(n)= c_0n^a+c_1n^{a-1}+\cdots$, we see that $H_X(n-d)= c_0(n-d)^a+c_1(n-d)^{a-1}+\cdots = c_0(n^a-adn^{a-1}+\cdots)+c_1n^{a-1}+\cdots$. So $H_X(n)-H_X(n-d)= c_0adn^{a-1}+\cdots$, giving that $\deg X\cap H = (a-1)!c_0ad=a!c_0d=d\cdot \deg X$, and we're done.